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QJEEM22Phc293: JEE Main 2022 Physics - Class 11 JEEM22Phc293 Solution: A steel wire of length and a copper wire of length , both of…

A steel wire of length 3.2 m(Ys=2.0×1011 Nm−2)3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right) and a copper wire of length 4.4 m(Yc=1.1×1011 Nm−2)4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right) , both of radius 1.4 mm1.4 \mathrm{~mm} are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm1.4 \mathrm{~mm} . The load applied, in Newton, will be: (\quad\left(\right. Given π=227\pi=\frac{22}{7} )

Multiple Choice Options

  • A 360
  • B 180
  • C 1080
  • D 154

Step-by-Step Verified Solution

Δls+Δlc=1.4\Delta {l_s} + \Delta {l_c} = 1.4

WlsYs×A+WlcYc×A=1.4×10−3{{W{l_s}} \over {{Y_s} \times A}} + {{W{l_c}} \over {{Y_c} \times A}} = 1.4 \times {10^{ - 3}}

W=1.4×10−3[3.22×(π×1.4×10−3)2+4.41.1×(π×1.4×10−3)2]110+11W = {{1.4 \times {{10}^{ - 3}}} \over {\left[ {{{3.2} \over {2 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}} + {{4.4} \over {1.1 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}}} \right]{1 \over {{{10}^{ + 11}}}}}}

W≃154 NW \simeq 154\,N

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